Respuesta :
Answer:
The limit that 97.5% of the data points will be above is $912.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 1500, \sigma = 300[/tex]
Find the limit that 97.5% of the data points will be above.
This is the value of X when Z has a pvalue of 1-0.975 = 0.025. So it is X when Z = -1.96.
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.96 = \frac{X - 1500}{300}[/tex]
[tex]X - 1500 = -1.96*300[/tex]
[tex]X = 912[/tex]
The limit that 97.5% of the data points will be above is $912.
Binomial distribution has mean 0 and standard deviation 1 and its z table helps with normal distributions. The needed limit is $912
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For the given case, let the data set's values given by considered statistical computer program is tracked by random variable X.
Then, by given data, we get:
[tex]X \sim N(1500, 300)[/tex]
and sample size = n = 200
We need value x such that
P(X > x) = 97.5% = 0.975 probability.
or
P(X ≤ x) = 1-0.975 = 0.025
Converting this to standard normal distribution (so that we can use z tables for ease), we get:
[tex]P(X \leq x) = P(Z \leq \dfrac{x -\mu}{\sigma}) = P(Z \leq \dfrac{x - 1500}{300}) = 0.025[/tex]
In the z tables available online, it is visible that the p value 0.025 is available for z = -1.96
Thus, we get:
[tex]z = \dfrac{x-1500}{300} = -1.96 \\\\or\\\\x = 300 \times -1.96 + 1500\\x= 912[/tex]
Thus, The needed limit is $912 after which there lies 97.5% of values of the considered data set.
Learn more about z scores here:
https://brainly.com/question/21118324
